My Math Forum rate of change/of a line

 Calculus Calculus Math Forum

 August 10th, 2016, 07:12 AM #1 Newbie   Joined: Aug 2016 From: Ireland Posts: 2 Thanks: 0 rate of change/of a line Hi I'm studying for a calculus test I understand how to get a slope and when a number is beside an x when written as an eqution y = mx + 1 the variable for x will be the slope but in my question the slope is x^2(x squared) which is totally confusing me,can anyone help walk me through how to get the answer here is the question If two variables x and y are related by the equation y = 3x^2 – x calculate the average rate of change of y with respect to x as x varies from 3 to 5. thanks,
 August 10th, 2016, 07:41 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 if $y = f(x) = 3x^2-x$, then the average rate of change of $f(x)$ is $\dfrac{f(5)-f(3)}{5-3}$
August 10th, 2016, 08:56 AM   #3
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,982
Thanks: 1575

note that the average rate of change is the slope of the $\color{red}{line}$ connecting the two points on the curve ... $(3,f(3))$ and $(5,f(5))$
Attached Images
 avg_roc.jpg (10.2 KB, 0 views)

 August 10th, 2016, 09:08 AM #4 Newbie   Joined: Aug 2016 From: Ireland Posts: 2 Thanks: 0 thanks for the reply Skeether much appreciated how come we sub in both 3 and 5 I thought they are both x values? I thought we could only sub in y and x to the slope formula? thanks,
 August 10th, 2016, 09:33 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 y is a function of x ... in other words, the value of y depends on the value of x at $x = 5$, $y = f(5) = 3(5)^2 - 5$ at $x = 3$, $y = f(3) = 3(3)^2 - 3$ slope = $\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{f(5)-f(3)}{5-3}$ Last edited by skeeter; August 10th, 2016 at 09:41 AM.

 Tags change or of, line, rate

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post puppypower123 Calculus 1 April 14th, 2016 05:48 PM Mr Davis 97 Pre-Calculus 2 May 5th, 2014 01:57 PM bongantedd Pre-Calculus 1 April 30th, 2014 12:40 AM henoshaile Calculus 1 October 21st, 2012 10:42 PM jeff Calculus 8 February 1st, 2012 01:30 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top